Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Heng Kan on October 02, 2018, 11:11:36 PM

For this question, we have to compute the limit of $f(z) = (z2)\logz2$ at z=2. As z approaches to $2$, $z2$ approaches to $0$ and $\logz2$ approaches to negative infinity. If this is a real function, we can use the L'Hopital's rule to compute the limit. So in this situation where it is a complex function, how do we find the limit? Thanks.

This is a tricky question, and most likely an erroneous, because $\log z2$ is a multivalued function and we have not defined the limits of such functions. And depending on how you make a definition, you'll get different answers. Observe that this is the only problem with $\log$.
Therefore I slightly change the function to $f(z)= (z2)\ln z2$ (here $z2$ is a real positive and $\ln $ is an ordinary logarithm). Try to write $(z2)$ in the polar form.

Here is a major hint: log in some contexts may actually mean ln when the variables is real. Also, note that the absolute value . maps from $\mathbb{C}\mapsto \mathbb{R}^+$. Hence, logz2 is just a real logarithm, and is in fact defined everywhere on the complex plane except for z=2. This can be done without using the deltaepsilon definition of limits (which is unfortunately hard to verify).
We have the expression $\lim\limits_{z \to 2} (z2) \lnz2$. The lim is of the form $0\cdot\infty$, meaning L'Hôpital's Rule applies.
Note that for L'Hôpital's Rule, $0 = \frac{1}{\infty}$. Hence, $(z2) = \frac{1}{(z2)^{1}}$, and the lim of both when they approach 2 are zero.
$\lim\limits_{z \to 2} (z2) \lnz2 = \lim\limits_{z \to 2}\frac{\lnz2}{(z2)^{1}}$, which is of the $\frac{\infty}{\infty}$ form, the desired indeterminate form.
We apply L'Hôpital's Rule: $\lim\limits_{z \to 2}\frac{\lnz2}{(z2)^{1}} \Rightarrow \lim\limits_{z \to 2}\frac{(z2)^{1}}{1(z2)^{2}}=\lim\limits_{z \to 2}\frac{(z2)^{1}}{(z2)^{2}}$. That looks very good, except the exponents are both negative, meaning that they switch places on the fraction bar.
We now arrive at $\lim\limits_{z \to 2}\frac{(z2)^{2}}{(z2)^{1}}$, and finally $\lim\limits_{z \to 2}(z2)$. We evaluate the limit directly: $(22) = 0$
∴ the lim of $(z2) \lnz2$ is zero.

It is fine, however we do not have L'Hôpital's Rule for complexvalued functions. However, if we want to prove that something goes to $0$, we actually need to prove that it's modules goes to $0$. So in fact we need to prove that $z2\ln z2$ goes to $0$ as $z\to 2$, which means that we trying to prove that $r\ln r\to 0$ as $r\to 0$.